# diff. eq.

Discussion in 'Homework Help' started by Hankhankhankhank, Oct 1, 2013.

1. ### Hankhankhankhank Thread Starter New Member

Sep 13, 2013
3
0
Will someone help me derive the diff. eq.? f(t) and any of its derivatives in terms of C1, R1, C2, R2, y(t), and any of y(t)'s derivatives. Thanks.

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2. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
You need to show YOUR best attempt to solve YOUR homework problem. That will provide the starting point for discussion to let us help steer you along the path to help YOU solve the problem.

3. ### Hankhankhankhank Thread Starter New Member

Sep 13, 2013
3
0
I tried to solve it, but I don't think I did it right. Please bear with me. Here is what I got:

f=y+1/C1*∫I*dt+I*R1

df=dy+1/C1*I+dI*R1

I=y/R2+C2*dy

dI=dy/R2+C2*d2y

df=dy+1/C1*(y/R2+C2*dy)+(dy/R2+C2*d2y)*R1

df=d2y*R1*C2+dy*(1+C2/C1+R1/R2)+y/(C1*R2)

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
The R2 and C2 form a current divider, i1=i2+i3.

5. ### Hankhankhankhank Thread Starter New Member

Sep 13, 2013
3
0
in my steps I my i1=i2+i3 was in the form:

I=y/R2+C2*dy

where i1=I, i2=y/R2, and i3=C2*dy

6. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
In general, try to make your work easy to follow by defining terms not on your diagram. For instant, in your first equation you use "I" but don't give any indication of what it is and require your readers to figure it out. Instead, make a notation that, "The current 'I' is flowing out of the supply (i.e., flowing left to right in R1)". Finally, the voltage across the capacitor C1 is equal to the definite integral from 0 to t, not the indefinite integral. Arguably, we should use a dummy variable for the integrand, but it isn't too confusing to just keep it as t.

Then, by convention, lower case variables are generally used for time varying signals and upper case variables for constants (and frequency domain variables, which we aren't dealing with here). It's good habit to stick to convention as much as reasonable.

Don't forget about the initial conditions. In this equation, C1 could have a voltage across it (positive on left side) at t=0, which we will call V1. So

$
f(t) \, = \, y(t) \, + \, \frac{1}{C_1}\int_0^t i(t) dt \, + \, V_1 \, + \, R_1 i(t)
$

Don't just use 'df' and 'dy'. These are infinitesimal quantities. Either use df/dt or use f' since the apostrophe is an accepted notation for the derivative with respect to a single variable, particularly time. So

$
\frac{df(t)}{dt} \, = \, \frac{dy(t)}{dt} \, + \, \frac{i(t)}{C_1} \, + \, R_1 \frac{di(t)}{dt}
$

or

$
f'(t) \, = \, y'(t) \, + \, \frac{1}{C_1}i(t) \, + \, R_1 i'(t)
$

Carrying the (t) in the notation is often dropped. This is, in part, reasonable because of the convention that lower case variables are understood to be varying functions of time. So

$
f' \, = \, y' \, + \, \frac{1}{C_1}i \, + \, R_1 i'
$

No problem here except the notation points already mentioned. So.

$
i \, = \, \frac{1}{R_2}y \, + \, C_2 y'
\
i' \, = \, \frac{1}{R_2} y' \, + \, C_2 y''
$

Still no problem. Cleaning up the notation, we have

$
f' \, = \, y' \, + \, \frac{1}{C_1} \left( \frac{1}{R_2}y \, + \, C_2 y' \right) \, + \, R_1 \left( \frac{1}{R_2} y' \, + \, C_2 y'' \right)
$

You are just fine. Again, cleaning up the notation.

$
f' \, = \, y'' R_1 C_2 \, + \, y' \left( 1 \, + \, \frac{C_2}{C_1}\, + \, \frac{R_1}{R_2} \right) \, + \, y \frac{1}{C_1 R_2}
$

At this point, a good sanity check is to ask if the units work out. The product of resistance times capacitance is time (known as an RC time constant). Each level of derivative results in the factor being divided by the unit of time. Finally, f(t) and y(t) are both voltages. So the left hand side of the equation has units of volts/time. Looking at each term in turn, we see that they also each have units of volts/time, so the units do work out.

anhnha and Hankhankhankhank like this.
7. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
So you can see how hard it can be for someone to back out what you are doing. If, instead, you indicate on your drawing I, I2, and I3 and then simply say

I = I2 + I3
I = y/R2 + C2*dy

It becomes very clear and easy to follow.

Remember, your reader doesn't have the mental picture you do and has to build it up based solely on what you provide them. Make it easy for them -- especially if your reader is also the person assigning your grade!

Hankhankhankhank likes this.